Chapter 19, Problem 19.114QP

Interpretation Introduction

Interpretation:

The pressure of hydrogen required to maintain the given reaction at equilibrium has to be calculated.

Concept Introduction:

• Nernst equation is one of the important equation in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

  ${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

  Where,

  ${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

  ${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

  $\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

  $\text{T}$ is the temperature

  $\text{n}$ is the number of electrons involved in a reaction

  $\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

  $\left[\text{Red}\right]$ is the concentration of the reduced species

  $\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({\text{25}}^{\text{°}}\text{C})$, after substituting the values of all the constants the equation can be written as

    ${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

• The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

• ${\text{p}}^{\text{H}}$ is the negative logarithm of the concentration of hydrogen ion in a solution.

${\text{p}}^{\text{H}}=-\log \left[{\text{H}}^{+}\right]$

  ${\text{p}}^{\text{OH}}$ is the negative logarithm of the concentration of hydroxide ion in the solution.

${\text{p}}^{\text{OH}}=-\log \left[{\text{OH}}^{-}\right]$

  For all aqueous solution at $\text{25}°\text{C}$ the ${\text{p}}^{\text{H}}$ and ${\text{p}}^{\text{OH}}$ related as given below,

${\text{p}}^{\text{H}}+{\text{p}}^{\text{OH}}=14.0$